path: root/source/know/concept/boussinesq-wave-theory/index.md

---
title: "Boussinesq wave theory"
sort_title: "Boussinesq wave theory"
date: 2023-01-07
categories:
- Physics
- Mathematics
- Fluid mechanics
- Fluid dynamics
layout: "concept"
---

In fluid mechanics, **Boussinesq wave theory**
consists of several equations to describe waves on a liquid's surface.
It was the first attempt to explain the nonlinear phenomenon of solitons,
which were not predicted by the linear theories existing at the time.



## Boundary conditions

Consider the [Euler equations](/know/concept/euler-equations/)
for an incompressible fluid with negligible viscosity:

$$\begin{aligned}
    \va{g} - \frac{\nabla p}{\rho}
    &= \pdv{\va{u}}{t} + (\va{u} \cdot \nabla) \va{u}
    \qquad \qquad
    \nabla \cdot \va{u}
    = 0
\end{aligned}$$

We rewrite the former using the velocity potential $$\Psi$$ for $$\va{u}$$
and the gravitational potential $$\Phi$$ for $$\va{g}$$,
such that $$\va{u} = \nabla \Psi$$ and $$\va{g} = - \nabla \Phi$$.
We also use a vector identity:

$$\begin{aligned}
    - \nabla \Phi - \frac{\nabla p}{\rho}
    &= \nabla \pdv{\Psi}{t} + \frac{1}{2} \nabla |\va{u}|^2 + (\nabla \cross \nabla \Psi) \cross \va{u}
\end{aligned}$$

Recall that the curl of a gradient is always zero, so the last term disappears.
Integrating in space yields integration constants $$C(t)$$ and $$p_0$$,
the latter representing atmospheric pressure:

$$\begin{aligned}
    \pdv{\Psi}{t} + \frac{1}{2} |\va{u}|^2
    &= -\Phi + \frac{p_0 - p}{\rho} + C
\end{aligned}$$

Consider a rectangular channel of depth $$h$$
extending infinitely far along the $$x$$-axis,
with a finite width along the $$y$$-axis.
We choose our coordinate system so that
$$z = 0$$ is the equilibrium water level,
and the bottom is at $$z = -h$$.
Let $$\eta(x, t)$$ be the deformation of the surface,
for which we want to find a wave equation.
All quantities are assumed to be constant in $$y$$,
so we only consider a 2D flow $$\va{u} = \big(u^{(x)}, u^{(z)}\big)$$.
At the surface $$z = \eta$$ we thus have:

$$\begin{aligned}
    \pdv{\Psi}{t} + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg)
    &= - g \eta + \frac{p_0 - p}{\rho} + C
\end{aligned}$$

Where $$g \eta = \Phi$$ with $$g \approx 9.81 \:\mathrm{m}/\mathrm{s}^2$$ on Earth.
Later, we will differentiate this formula in $$x$$,
so we can already set $$C = 0$$ now, since it will vanish then anyway.
Furthermore, we assume that at the surface $$z = \eta$$
the pressures are in equilibrium $$p_0 = p$$, leaving:

$$\begin{aligned}
    \boxed{
        \pdv{\Psi}{t} + \frac{1}{2} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg) + g \eta
        = 0
    }
\end{aligned}$$

This is called the **free surface boundary condition**.
Obviously, if $$z = \eta$$, then $$\eta \!-\! z = 0$$.
Taking the [material derivative](/know/concept/material-derivative/)
of this fact gives the following relation:

$$\begin{aligned}
    0
    = \frac{\mathrm{D}}{\mathrm{D} t} (\eta - z)
    = \pdv{\eta}{t} + \pdv{z}{t} + \va{u} \cdot \nabla \eta - \va{u} \cdot \nabla z
\end{aligned}$$

Since $$\eta$$ only depends on $$x$$ and $$t$$,
this becomes the **kinematic boundary condition**:

$$\begin{aligned}
    \boxed{
        \pdv{\eta}{t} + u^{(x)} \pdv{\eta}{x} - u^{(z)}
        = 0
    }
\end{aligned}$$

The equations will be derived from these two fundamental boundary conditions.



## Boussinesq approximation

Let us take a Taylor expansion of the velocity potential $$\Psi(x, z, t)$$
at the bottom $$z = -h$$:

$$\begin{aligned}
    \Psi(x, z)
    = \Psi(x, -h) + (z + h) \: \Psi_z(x, -h) + \frac{(z + h)^2}{2} \: \Psi_{zz}(x, -h) + ...
\end{aligned}$$

Because the fluid is incompressible, this can be rewritten.
Laplace's equation tells us:

$$\begin{aligned}
    \nabla \cdot \va{u}
    = \nabla^2 \Psi
    = 0
    \qquad \implies \qquad
    \:\:\:\Psi_{zz}
    = - \Psi_{xx}
\end{aligned}$$

Which we use for the expansion's second-order term.
Similarly, for the fourth-order term:

$$\begin{aligned}
    0
    = \nabla^2 \nabla^2 \Psi
    &= \Psi_{zzzz} + \pdvn{2}{}{z} \Psi_{xx} + \pdvn{2}{}{x} \Psi_{zz} + \Psi_{xxxx}
    \\
    &= \Psi_{zzzz} + \pdvn{2}{}{x} \Psi_{zz} - \pdvn{2}{}{x} \Psi_{xx} + \Psi_{xxxx}
    \\
    &= \Psi_{zzzz} - \Psi_{xxxx}
\end{aligned}$$

And so on for higher orders.
In the Taylor expansion, all even derivatives can be rewritten in this way,
and all odd derivatives can be split into a single $$\ipdv{}{z}$$ and an even $$x$$-derivative:

$$\begin{aligned}
    \Psi(x, z)
    &= \Psi + (z \!+\! h) \Psi_z - \frac{(z \!+\! h)^2}{2} \Psi_{xx}
    - \frac{(z \!+\! h)^3}{6} \pdv{}{z} \Psi_{xx} + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx} + ...
    \\
    &= \bigg(\! \Psi - \frac{(z \!+\! h)^2}{2} \Psi_{xx} + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx} - ... \bigg)
    \!+\! \bigg(\! (z \!+\! h) \Psi_z - \frac{(z \!+\! h)^3}{6} \pdvn{2}{}{x} \Psi_{z} + ... \bigg)
\end{aligned}$$

By definition $$\Psi_z = u^{(z)}$$,
but the bottom is solid, so at $$z = -h$$ we need $$\Psi_z = 0$$,
leaving:

$$\begin{aligned}
    \boxed{
        \Psi(x, z)
        = \Psi(x, -h) - \frac{(z \!+\! h)^2}{2} \Psi_{xx}(x, - h) + \frac{(z \!+\! h)^4}{24} \Psi_{xxxx}(x, - h) - ...
    }
\end{aligned}$$

This result is exact for an inviscid incompressible fluid,
but once the Taylor series is truncated at a finite number of terms,
this is known as the **Boussinesq approximation**.
In effect, this removes all $$z$$-derivatives from the problem,
and will enable us to describe the surface dynamics
based on $$\Psi$$'s behaviour near the channel's bottom.

This expression for $$\Psi$$ gives the flow components,
where we define $$f(x, t) \equiv \Psi_x(x, -h, t)$$
as the value of $$u^{(x)}$$ at the bottom:

$$\begin{aligned}
    u^{(x)}(x, z)
    = \pdv{\Psi}{x}
    &= f(x) - \frac{(z \!+\! h)^2}{2} f_{xx}(x) + \frac{(z \!+\! h)^4}{24} f_{xxxx}(x) - ...
    \\
    u^{(z)}(x, z)
    = \pdv{\Psi}{z}
    &= - (z \!+\! h) f_x(x) + \frac{(z \!+\! h)^3}{6} f_{xxx}(x) - ...
\end{aligned}$$

The Boussinesq approximation is the basis of many other shallow-water wave theories,
most notably the [Korteweg-de Vries equation](/know/concept/korteweg-de-vries-equation/).



## Two coupled equations

Inserting this result (without truncating)
into the kinematic boundary condition with $$z = \eta$$:

$$\begin{aligned}
    0
    &= \pdv{\eta}{t} + \pdv{\eta}{x} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
\end{aligned}$$

And into the free surface boundary condition after differentiating it with respect to $$x$$:

$$\begin{aligned}
    0
    &= \pdv{u^{(x)}}{t} + \frac{1}{2} \pdv{}{x} \bigg( \big(u^{(x)}\big)^2 + \big(u^{(z)}\big)^2 \bigg) + g \pdv{\eta}{x}
    \\
    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
    + \frac{1}{2} \pdv{}{x} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)^2 \\
    &\qquad + \frac{1}{2} \pdv{}{x} \bigg( \!-\! (\eta \!+\! h) f_x + \frac{(\eta \!+\! h)^3}{6} f_{xxx} - ... \bigg)^2 + g \pdv{\eta}{x}
    \\
    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 - f f_{xx}) + ... \bigg) + g \pdv{\eta}{x}
\end{aligned}$$

Switching to a shorter notation for derivatives,
we now have the following set of equations:

$$\begin{aligned}
    0
    &= \eta_t + \eta_x \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
    + \bigg( (\eta \!+\! h) f_{x} - \frac{(\eta \!+\! h)^3}{6} f_{xxx} + ... \bigg)
    \\
    0
    &= \pdv{}{t} \bigg( f - \frac{(\eta \!+\! h)^2}{2} f_{xx} + ... \bigg)
    + \frac{1}{2} \pdv{}{x} \bigg( f^2 + (\eta \!+\! h)^2 (f_x^2 - f f_{xx}) + ... \bigg) + g \eta_x
\end{aligned}$$

Now we must decide which terms to keep, i.e. where to truncate the expansion.
Let us therefore introduce the characteristic length scales $$\eta \sim a$$ and $$x \sim \lambda$$,
and assume that the water is shallow compared to the waves' length ($$\lambda \gg h$$ by a lot),
and that the waves' amplitude is small compared to the channel's depth ($$h \gg a$$).
Specifically, we assume:

$$\begin{aligned}
    \frac{a}{h}
    \gg \frac{h}{\lambda}
    \gg \frac{a}{\lambda}
\end{aligned}$$

We also introduce characteristic horizontal velocity scales $$u_0$$ and $$f_0$$
respectively at the surface and at the bottom.
Finally, let there be a characteristic time scale $$\lambda / u_0$$ for the surface dynamics.
Inserting all these scales into the two boundary conditions yields:

$$\begin{aligned}
    0
    &\sim \frac{a u_0}{\lambda}
    + \frac{a}{\lambda} \bigg( f_0 - \frac{(a \!+\! h)^2}{2 \lambda^2} f_0 + ... \bigg)
    + \bigg( \frac{(a \!+\! h)}{\lambda} f_0 - \frac{(a \!+\! h)^3}{6 \lambda^3} f_0 + ... \bigg)
    \\
    0
    &\sim \frac{u_0}{\lambda} \bigg( f_0 - \frac{(a \!+\! h)^2}{2 \lambda^2} f_0 + ... \bigg)
    + \frac{1}{2 \lambda} \bigg( f_0^2 + \frac{(a \!+\! h)^2}{\lambda^2} f_0^2 + ... \bigg)
    + \frac{g a}{\lambda}
\end{aligned}$$

Intuitively, we expect that $$u_0 \gg f_0$$, and this is indeed true:
from a linearization of this problem (given in the next section),
it turns out that $$f_0 / u_0 \sim a / h$$ and $$u_0 \approx \sqrt{g h}$$.
Multiplying the former equation by $$1 / u_0$$
and the latter by $$\lambda / u_0^2$$ this leads to:

$$\begin{aligned}
    0
    &\sim \frac{a}{\lambda}
    + \frac{a}{\lambda} \bigg( \frac{a}{h} - \frac{(a \!+\! h)^2}{2 \lambda^2} \frac{a}{h} + ... \bigg)
    + \bigg( \frac{(a \!+\! h)}{\lambda} \frac{a}{h} - \frac{(a \!+\! h)^3}{6 \lambda^3} \frac{a}{h} + ... \bigg)
    \\
    0
    &\sim \bigg( \frac{a}{h} - \frac{(a \!+\! h)^2}{2 \lambda^2} \frac{a}{h} + ... \bigg)
    + \frac{1}{2} \bigg( \frac{a^2}{h^2} + \frac{(a \!+\! h)^2}{\lambda^2} \frac{a^2}{h^2} + ... \bigg)
    + \frac{a}{h}
\end{aligned}$$

The smallest term we will include is $$a h^2 / \lambda^3$$;
anything smaller (specifically containing $$a^2 / \lambda^2$$) will be discarded.
Of course, this decision is arbitrary:
higher-order approximations exist for deeper water and/or taller waves,
but we stick with Boussinesq's original choice, leaving:

$$\begin{aligned}
    0
    &= \eta_t + \eta_x f + (\eta \!+\! h) f_{x} - \frac{h^3}{6} f_{xxx}
    \\
    0
    &= \pdv{}{t} \bigg( f - \frac{h^2}{2} f_{xx} \bigg)
    + \frac{1}{2} \pdv{}{x} \big( f^2 \big) + g \eta_x
\end{aligned}$$

Rearranging this gives the **Boussinesq equations**
for nonlinear waves on shallow water:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \pdv{\eta}{t} + \pdv{}{x} \Big( (\eta \!+\! h) f \Big)
            &= \frac{h^3}{6} \pdvn{3}{f}{x}
            \\
            \pdv{f}{t} + f \pdv{f}{x} + g \pdv{\eta}{x}
            &= \frac{h^2}{2} \pdv{}{t} \pdvn{2}{f}{x}
        \end{aligned}
    }
\end{aligned}$$

If we instead took $$\lambda$$ to be even larger compared to $$h$$,
the right-hand sides of these equations would have vanished,
yielding a form of the so-called *shallow water equations*.



## Single equation

We would like to combine these two equations into a single one for $$\eta$$,
but their nonlinear nature makes it hard to eliminate $$f$$ directly.
To get around this, Boussinesq opted to make a lower-order version of these equations,
to use as a guide for some additional approximations
to help handle the higher-order version.

In the above discussion of the terms' relative sizes,
let us instead choose $$a / \lambda$$ as the highest order to include,
thereby reducing the equations to:

$$\begin{aligned}
    \eta_t + h f_x
    = 0
    \qquad \qquad
    f_t + g \eta_x
    = 0
\end{aligned}$$

Respectively differentiating them with respect to $$t$$ and $$x$$,
and then substituting $$f_{xt}$$ in the former using the latter,
we get Lagrange's linear wave equation:

$$\begin{aligned}
    \pdvn{2}{\eta}{t} - g h \pdvn{2}{\eta}{x}
    = 0
\end{aligned}$$

It is well-known that such a problem has a general solution
consisting of an arbitrary forward-moving part $$\eta_{+}$$
and a backward-moving part $$\eta_{-}$$,
both going at a constant velocity $$\sqrt{g h}$$,
and neither of which change shape over time:

$$\begin{aligned}
    \eta(x, t)
    = \eta_{+}\big(x - \sqrt{g h} t\big) + \eta_{-}\big(x + \sqrt{g h} t\big)
\end{aligned}$$

Let us consider only forward-moving waves $$\eta(x \!-\! \sqrt{g h} t)$$,
such that we can rewrite $$t$$-derivatives as $$x$$-derivatives with a factor $$-\sqrt{g h}$$.
Our linearized free-surface equation thus becomes:

$$\begin{aligned}
    0
    = \pdv{\eta}{t} + h \pdv{f}{x}
    = - \sqrt{g h} \pdv{\eta}{x} + h \pdv{f}{x}
    \qquad \implies \qquad
    f
    = \sqrt{\frac{g}{h}} \eta + C
\end{aligned}$$

The integration constant $$C$$ can be removed by absorbing it into $$\eta$$.
Effectively, we have seen that, at least as a first-order approximation,
$$\eta$$ is proportional to $$f$$.
Note that this analysis justifies our earlier assumption
that $$f_0 / u_0 \sim a / h$$ and $$u_0 \approx \sqrt{g h}$$.

Armed with this knowledge, we return
to the higher-order equations after some rearranging:

$$\begin{aligned}
    \eta_{t} + h f_x + \pdv{}{x} \bigg( \eta f - \frac{h^3}{6} f_{xx} \bigg)
    &= 0
    \\
    f_{t} + g \eta_{x} + \frac{1}{2} \pdv{}{x} \bigg( f^2 - h^2 f_{xt} \bigg)
    &= 0
\end{aligned}$$

Inserting $$f = \sqrt{g / h} \eta$$ into both equations
and multiplying the latter by $$h$$ yields:

$$\begin{aligned}
    \eta_{t} + \sqrt{g h} \eta_x + \sqrt{\frac{g}{h}} \pdv{}{x} \bigg( \eta^2 - \frac{h^3}{6} \eta_{xx} \bigg)
    &= 0
    \\
    \sqrt{g h} \eta_{t} + g h \eta_{x} + \frac{1}{2} \pdv{}{x} \bigg( g \eta^2 - h^2 \sqrt{g h} \eta_{xt} \bigg)
    &= 0
\end{aligned}$$

Respectively differentiating by $$t$$ and $$x$$
and assuming a travelling wave $$\eta(x \!-\! \sqrt{g h} t)$$
such that we can rewrite $$\ipdv{}{t} = - \sqrt{g h} \ipdv{}{x}$$:

$$\begin{aligned}
    \eta_{tt} + \sqrt{g h} \eta_{xt} - g \pdvn{2}{}{x} \bigg( \eta^2 - \frac{h^3}{6} \eta_{xx} \bigg)
    &= 0
    \\
    \sqrt{g h} \eta_{tx} + g h \eta_{xx} + \frac{g}{2} \pdvn{2}{}{x} \bigg( \eta^2 + h^3 \eta_{xx} \bigg)
    &= 0
\end{aligned}$$

Subtracting the latter from the former yields the following equation containing only $$\eta$$:

$$\begin{aligned}
    \eta_{tt} + \sqrt{g h} \eta_{xt} - \sqrt{g h} \eta_{tx} - g h \eta_{xx}
    - \pdvn{2}{}{x} \bigg( g \Big( \eta^2 - \frac{h^3}{6} \eta_{xx} \Big) + \frac{g}{2} \Big( \eta^2 + h^3 \eta_{xx} \Big) \bigg)
    &= 0
\end{aligned}$$

After cleaning up, this becomes the **Boussinesq equation** for the shape of travelling waves:

$$\begin{aligned}
    \boxed{
        \pdvn{2}{\eta}{t} - g h \pdvn{2}{\eta}{x} - g h \pdvn{2}{}{x} \bigg( \frac{3}{2} \frac{\eta^2}{h} + \frac{h^2}{3} \pdvn{2}{\eta}{x} \bigg)
        = 0
    }
\end{aligned}$$

Clearly, the assumption of non-deforming waves $$\eta(x \!-\! \sqrt{g h} t)$$
was essential to get this equation.
But what if solving it yields a wave without that property?
Can it be trusted?
Fortunately yes: the first two terms ($$\eta_{tt}$$ and $$g h \eta_{xx}$$),
were not affected by that assumption (this is easy to see),
and the others are small according to the characteristic scales:

$$\begin{aligned}
    0
    \sim \frac{a u_0^2}{\lambda^2} - g h \frac{a}{\lambda^2}
    - g h \frac{1}{\lambda^2} \bigg( \frac{3}{2} \frac{a^2}{h} + \frac{h^2}{3} \frac{a}{\lambda^2} \bigg)
\end{aligned}$$

After dividing out $$a / \lambda$$,
we see that the last two terms are roughly $$a / \lambda$$ and $$h^3 / \lambda^3$$,
meaning they are much smaller than the first two,
which are both on the order of $$h / \lambda$$.



## References
1.  J. Boussinesq,
    [Théorie des ondes et des remous qui se propagent le long d'un canal rectangulaire horizontal, en communiquant au liquide contenu dans ce canal des vitesses sensiblement pareilles de la surface au fond](http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16416&Deb=63&Fin=116&E=PDF),
    1872, Bibliothèque nationale de France.
2.  E.M. de Jager,
    [On the origin of the Korteweg-de Vries equation](https://arxiv.org/abs/math/0602661),
    University of Amsterdam.
3.  D. Dutykh, F. Dias,
    [Dissipative Boussinesq equations](https://doi.org/10.1016/j.crme.2007.08.003),
    2007, Elsevier.